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ID: 1002932 • Letter: H
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home / study / science / physics / questions and answers / a living specimen in equilibrium with the atmosphere ... Your question has been answered! Rate it below. Let us know if you got a helpful answer. Question A living specimen in equilibrium with the atmosphere contains one atom of 14C (half-life = 5730 yr) for every 7.70 *10^11 stable carbon atoms. A bone fragment contains 1.00*10^21 carbon nuclei. When the sample is placed inside a shielded beta counter with 75.0% counting efficiency, 82,000 counts are accumulated in one year. Assuming that the cosmic-ray flux and the Earth's atmosphere have not changed appreciably since the sample was formed, find the age of the sample. I know the answer is 2993 years, but don't know why. Please explain! THANK YOU!
Explanation / Answer
Let 't' be the age of the bone fragment
rate constant, k = 0.693 / t1/2 = 0.693 / 5730 yr = 1.21x10-4 yr-1
Initial number of atms of C-14 in the organic sample, N0
= (1 C-14 / 7.70x1011 C-12) x 1.00x1021 C12
=> N0 = 1.30x109 C-14 atoms
The radioactive decay of 1 C-14 releases 1 beta - particle.
Total number of beta-particle accumulated in 1 year = 82000 x 100/75 = 109333 beta - particles
Hence the total number of C-14 atms decayed in 1 year( 't' th year) = 109333 C-14 atoms
If Nt - 1 is the number of C-14 atoms after (t-1) year and Nt is the number of C-14 atoms after t year
=> Nt - 1 - Nt = 109333 C-14 atoms
=> Nt - 1 = (109333 + Nt) C-14 atms
kt = ln(N0 / Nt )
=> 1.21x10-4 yr-1 x 1 yr = ln(Nt - 1 / Nt ) -------- (2)
=> 1.21x10-4 yr-1 x 1 yr = ln (109333 + Nt) / Nt
=> (109333 + Nt) / Nt = (exp)1.21x10-4 = 1.000121
=> 109333 + Nt = 1.000121 x Nt
=> Nt = 9.0395x108 C-14 atoms
Also
kt = ln(N0 / Nt )
=> 1.21x10-4 yr-1 x t = ln(1.30x109 / 9.0395x108) -------- (3)
=> t = ln(1.30x109 / 9.0395x108) / 1.21x10-4 yr-1 = 2993 years (answer)
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